Erm - I think a better solution would be split the initial quadrilateral into thirds - this also allows correction of the initial scale edges, --------- | | | | --------- | | | | --------- | | | | --------- On 6 Jan 2013, at 18:31, Marc LeBrun wrote:
="Henry Baker" <hbaker1@pipeline.com>
Can a smooth surface -- e.g., a hemisphere -- be approximated by a convex quadrangulated surface?
The answer is trivially yes -- the approximation is a cube -- but is there a way to quadrangulate it with smaller & smaller quadrilaterals to get a better & better approximation?
The answer is, again, trivially yes. Just keep stacking smaller aligned cubes into any gaps. Stop whenever your approximation is close enough. You will have a nice "cubic crystallized" model not only at the surface, but of the volume (that you can readily feed into your 3D printer if you want<;-).
Also keep in mind that, whereas any 3 points on a surface always define a planar triangle, any 4 points on a surface need not be co-planar; consider a saddle for example. So incremental triangularization schemes are inherently easier and more robust than successive quadrangularizations will be.
And what exactly does "smooth" buy us? Let's just ask instead: how would we quadrangulate a rounded tetrahedron?
Although the intent seems intuitive, if all sorts of cases must be patched with a stream of new rules then I'd guess that it's probably because the problem to be solved isn't sufficiently precisely articulated yet.
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